CPMagical CP Library

Problem solution · Python

ABC401 D — Logical Filling

ABC401 D — Logical Filling: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Sliding window or two pointers
Source
KATO-Hiro AtCoder Solutions
Length
88 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For ABC401 D — Logical Filling, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 88 lines of Python from the credited upstream file abc401_d.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from KATO-Hiro AtCoder Solutions by KATO-Hiro and is used under the CC0-1.0 licence.

Open problem ↗Original source ↗
Full codeABC401 D — Logical Filling · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*- from typing import List  def run_length_encoding(iterable: list) -> List[list]:    """    Args:        iterable: A list of numbers or strings.     Returns:        A list containing consecutive characters and their count.     See:    https://qiita.com/DaikiSuyama/items/07e237b7372e7c7b3432    """     from itertools import groupby     results = [[key, len(list(group))] for key, group in groupby(iterable)]     return results  def main():    import sys     input = sys.stdin.readline     n, k = map(int, input().split())    s = list(input().rstrip())     # 前処理: o が連続しない => o の両隣は . と言い換え    for i, si in enumerate(s):        if si == "o":            if i - 1 >= 0:                s[i - 1] = "."            if i + 1 < n:                s[i + 1] = "."     # o.???.o.?????のような構造    # o の数を数えて、残りを?に割り当てる    remain = k - s.count("o")     # ??? の区間と取りうる最大の o の個数を求める    results = run_length_encoding(s)    total = 0    ps = list()     for key, count in results:        if key == "?":            ps.append((total, total + count))  # [left, right)         total += count     candidate_max = 0     for left, right in ps:        candidate_max += (right - left + 1) // 2     # 場合分け    # 1. o が 既に k 個使われている: ? を 全て . に置き換え    if remain == 0:        for i, si in enumerate(s):            if si == "?":                s[i] = "."    # 2. ???に埋められる上限まで必要    # 2.1 偶奇で場合分け: 偶数の場合は ?、奇数の場合は o.o.oで埋める    elif remain == candidate_max:        for left, right in ps:            if (right - left) % 2 == 0:                continue             for i in range(right - left):                if i % 2 == 0:                    s[left + i] = "o"                else:                    s[left + i] = "."    # 3. それ以外    else:        pass     print("".join(s))  if __name__ == "__main__":    main()

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗